
In the figure, \[{S_1}\]and \[{S_2}\]are identical springs. The oscillation frequency of the mass m is\[f\]. If one spring is removed, the frequency will become

Fig: The springs attached to the mass m
A. \[f\]
B. \[f \times 2\]
C. \[f \times \sqrt 2 \]
D. \[\dfrac{f}{{\sqrt 2 }}\]
Answer
161.1k+ views
Hint:We use the frequency of the spring-block system by finding the equivalent of the springs connected to the block. The frequency of the spring-block system is proportional to the square root of the spring constant.
Formula used:
\[\dfrac{1}{{{k_{eq}}}} = \dfrac{1}{{{k_1}}} + \dfrac{1}{{{k_2}}} + \dfrac{1}{{{k_3}}} \ldots + \dfrac{1}{{{k_n}}}\], here \[{k_{eq}}\]is the equivalent spring constant when the springs are connected in series.
\[{k_{eq}} = {k_1} + {k_2} + {k_3} \ldots + {k_n}\], here \[{k_{eq}}\] is the equivalent spring constant when the springs are connected in parallel.
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{{K_{eq}}}}{m}} \],
where f is the frequency of spring-mass system.
Complete step by step solution:

Fig: The springs attached to the mass m
As the springs are identical, so the spring constant of both springs will be equal.
Let the spring constant of the given two springs is k.
Both springs are connected in series. So, the equivalent spring constant is,
\[{k_{eq}} = k + k\]
\[{k_{eq}} = 2k\]
So, the frequency of the initial spring-block system is,
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{2k}}{m}} \]
When one of the spring is removed then the net spring constant of the system will be k,
So, the frequency of the final spring-block system is,
\[{f_2} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \]
On dividing both the equations, we get
\[\dfrac{f}{{{f_2}}} = \dfrac{{\dfrac{1}{{2\pi }}\sqrt {\dfrac{{2k}}{m}} }}{{\dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} }}\]
\[\dfrac{f}{{{f_2}}} = \sqrt {\dfrac{{2k}}{k}} \]
\[{f_2} = \dfrac{f}{{\sqrt 2 }}\]
So, the frequency of the spring-block system when one of the spring is removed will be \[\dfrac{f}{{\sqrt 2 }}\]
Therefore, the correct option is (D).
Note:We should be careful while finding the equivalent spring constant of the combination of the springs. When the compression in all the springs are same when block is displaced, then the springs are in parallel combination.
Formula used:
\[\dfrac{1}{{{k_{eq}}}} = \dfrac{1}{{{k_1}}} + \dfrac{1}{{{k_2}}} + \dfrac{1}{{{k_3}}} \ldots + \dfrac{1}{{{k_n}}}\], here \[{k_{eq}}\]is the equivalent spring constant when the springs are connected in series.
\[{k_{eq}} = {k_1} + {k_2} + {k_3} \ldots + {k_n}\], here \[{k_{eq}}\] is the equivalent spring constant when the springs are connected in parallel.
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{{K_{eq}}}}{m}} \],
where f is the frequency of spring-mass system.
Complete step by step solution:

Fig: The springs attached to the mass m
As the springs are identical, so the spring constant of both springs will be equal.
Let the spring constant of the given two springs is k.
Both springs are connected in series. So, the equivalent spring constant is,
\[{k_{eq}} = k + k\]
\[{k_{eq}} = 2k\]
So, the frequency of the initial spring-block system is,
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{2k}}{m}} \]
When one of the spring is removed then the net spring constant of the system will be k,
So, the frequency of the final spring-block system is,
\[{f_2} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \]
On dividing both the equations, we get
\[\dfrac{f}{{{f_2}}} = \dfrac{{\dfrac{1}{{2\pi }}\sqrt {\dfrac{{2k}}{m}} }}{{\dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} }}\]
\[\dfrac{f}{{{f_2}}} = \sqrt {\dfrac{{2k}}{k}} \]
\[{f_2} = \dfrac{f}{{\sqrt 2 }}\]
So, the frequency of the spring-block system when one of the spring is removed will be \[\dfrac{f}{{\sqrt 2 }}\]
Therefore, the correct option is (D).
Note:We should be careful while finding the equivalent spring constant of the combination of the springs. When the compression in all the springs are same when block is displaced, then the springs are in parallel combination.
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
